A227693 Integer nearest to (F[2n+1](S(n)))^2, where F[2n+1](x) are Fibonacci polynomials of odd indices [2n+1] and S(n) = Sum_{i=0..2} (C(i)*(log(log(A*(B+n^2))))^(2i)) (see coefficients A, B, C(i) in comments).

4, 25, 168, 1229, 9595, 78527, 664408, 5759130, 50833725, 455019102, 4118498801, 37616575907, 346165453783, 3205869110911, 29851888456753, 279286334215803, 2623780688311969, 24739953477533166, 234041108830344356, 2220562531262307905, 21124612016460745383, 201448482556532026684, 1925296277838503159171, 18437832696789559015711, 176901280909820032014422

Offset: 1

Author

Vladimir Pletser, Jul 19 2013

Keywords

• nonn

Comments

Coefficients are A= 0.1641239, B= 10.0861, C(0)=0 .9976796712309498, C(1)= 7.445960495e-02, C(2)= -6.73751166802e-03.

This sequence gives a good approximation of pi(10^n) (A006880); see (A227694).

To obtain this sequence, remark first that the square root of the first values of pi(10^n) (A006880) (see (A221205)) are close to odd indices Fibonacci numbers F[2n+1](1). Switching to odd indices Fibonacci polynomials F[2n+1](x), one obtains the sequence a(n) by computing x as a function of n such that (F[2n+1](x))^2 fit the values of pi(10^n) for 1<=n<=25.

Examples

			For n =1, F[3](x) = x^2+1; replace x by Sum_{i=0..2} (C(i)*(log(log(A*(B+1))))^(2i))= 1.016825… to obtain a(1)= round((F[3]( 1.016825…))^2)=4.
		

Crossrefs

Cf. A006880, A227694.

Programs

Maple

with(combinat, fibonacci): A:= 0.1641239: B:= 10.0861: C(0):= .9976796712309498: C(1):=7.445960495E-02: C(2):= -6.73751166802E-03: b:=n->log(log(A*(B+n^2))): c:=n->sum(C(i)*(b(n))^(2*i), i=0..2): seq(round(fibonacci(2*n+1, c(n))^2), n=1..25);

Formula

a(n) = round((F[2n+1](Sum_{i=0..2} (C(i)*(log(log(A*(B+n^2))))^(2i))))^2).